Trigonometry and Modelling Mixed Exercise a i ii sin40 cos0 - cos40 sin0 sin(40-0 ) sin0 cos - sin cos 4 cos - sin 4 sin As cos(x - y) sin y cos xcos y + sin xsin y sin y () Draw a right-angled triangle, where sin x cos(4 + ) cos60 iii - tan + tan tan4 - tan + tan4 tan tan(4 - ) tan0 Using Pythagoras' theorem, a - 4 Þ a So cos x Sustitute into () : cos y + sin y sin y Þ cos y + sin y sin y Þ cos y sin y - Þ Þ tan y sin y ö tan y tan y - cos y ( +) ( -)( +) 4 + + a tan A, tan B since y x - The angle required is (A- B). tan A - tan B Using tan( A- B) + tan Atan B Þ A- B 4 - + Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
4 6 a sin A,cos A 4 sin B,cos B Using sin B sinc c sin(q - 0 ) sin(q + 0 ) Þ 4 Þ sin(q - 0 ) 4sin(q + 0 ) Þ (sinq cos0 - cosq sin0 ) 4(sinq cos0 + cosq sin0 ) Þ sinq cos0 9cosq sin0 Þ sinq cosq 9 sin0 cos0 9tan0 Þ tanq 9 As the three values are consecutive terms of an arithmetic progression, sin(q - 0 ) - cosq sinq - sin(q - 0 ) Þ sin(q - 0 ) sinq + cosq Þ (sinq cos0 - cosq sin0 ) sinq + cosq Þ sinq - cosq sinq + Þ sinq - cosq ( +) Þ tanq + - cosq Calculator value is q tan - + - 7 No other values as q is acute. 7 a i ii sin( A+ B) sin Acos B + cos Asin B + 4 6 6 tanb - 6 9 44 6 44 9 0 9 - tan B - tan B cosc cos( 80 - ( A+ B) ) -cos( A+ B) -(cos Acos B - sin Asin B) - 4 - ö - 6 cosx - sin x ö cosy cos y - ö 0-8 - - 9 ö 0-4 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
7 c tanq, for 0 q p q p 6, 7p 6 Þq p, 7p 8 a c i ii tan(x + y) + - 7 7 tan x - tan y tan(x - y) + tan x tan y As x and y are acute, and x > y, x - y is acute So x - y p 4 it cannot e p ö 4 + 6 sin(x - y) (sin xcos y - cos xsin y) sin xcos y cos xsin y Þ tan x tan y so tan x tan y k 4-9k tan x + tan y - tan x tan y sin(x + y) sin xcos y + cos xsin y tanx - ö 6 6 k tan x - tan x k - 9 4 k 9 a sinq + sin q sinq - sin q cosq sinq cosq Þ tanq 0 a cos sin Þ cosq - sinq 0 Þ- sin q - sinq 0 Þ sin q + sinq - 0 a, and c sin sin 0 Using the quadratic formula sinq - ± - 4()(-) () - ± 4 sinq 0.86, for -p q p sinq is positive so solutions in the first and second quadrants q sin - 0.86, p - sin - 0.86 q 0.87,.94 ( d.p.) a cos(x - 60 ) cos xcos60 + sin xsin60 cos x sin x So sin x cos x + sin x Þ - ö sin x cos x Þ tan x - 4-4 - tan x 0.44 ( d.p.), in the 4 - interval 0 q 60 tanq is positive so solutions in the first and third quadrants x.8, 0.8 ( d.p.) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
a cos(x + 0 ) sin(90-0 - x) sin(70 - x) sin70 cos x - cos70 sinx () 4sin(70 + x) 4sin70 cos x + 4cos70 sin x () As () () 4sin70 cos x + 4cos70 sin x sin70 cos x - cos70 sinx sin xcos70 -sin70 cos x tan x - tan70 tan x - tan70, for 0 q 80 tanq is negative so the solution is in the second quadrant x 80 + tan - - tan70 ö x 80 - tan - (-.648) x 80 - (-8.8 ). ( d.p.) a Draw a right-angled triangle and find sina and cosa. Þ sina, cosa 4 sin(q + a ) + 4cos(q + a ) (sinq cosa + cosq sina ) + 4(cosq cosa - sinq sina ) 4 sinq + cosq ö + 4 4 cosq - sinq ö cos(x + 70 ) cos x cos70 - sin x sin70 (-0.8)(0) - (0.6)(-) 0 + 0.6 0.6 cos(x + 40 ) cos x cos40 - sin x sin40 (-0.8)(-) - (0.6)(0) 0.8-0 0.8 4 a One example is sufficient to disprove a statement. Let A 60, B 0 sec( A + B) sec(60 + 0 ) sec60 cos60 sec A sec60 cos60 sec B sec0 cos0 So sec A + sec B + So sec(60 + 0 ) ¹ sec60 + sec0 Þ sin( A + B) sec A + sec B is not true for all values of A, B. LHS tanq + cotq sinq cosq + cosq sinq sin q + cos q sinq cosq sinq Using sin q + cos q, and sinq sinq cosq So LHS sinq cosecq RHS sinq + 9 cosq + 6 cosq - sinq cosq cosq Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4
a Using tanq tanq - tan q with q p 8 Þ tan p 4 tan p 8 - tan p 8 Sketch y sin(x - 60 ) y first translating y sin x y 60 to the right and then stretching the result in the y direction y scale factor. Let t tan p 8 So t - t Þ - t t Þ t + t - 0 Þ t - ± 8 -± As p 8 is acute, tan p 8 - ± is positive, so tan p 8 - tan p 8 tan p 4 + p ö 8 tan p + tan p 4 8 - tan p tan p 4 8 - ( + ) ( - ) ( + ) + + - - - + 6 a Let sin x - cos x Rsin(x -a) Rsin xcosa - Rcos xsina R > 0, 0 < a < 90 Compare sin x: Rcosa () Compare cos x: Rsina () Divide () y (): tana Þ a 60 R + 4 Þ R So sin x - cos x sin(x - 60 ) Graph meets y-axis when x 0, i.e. y sin(-60 ) -, at 0,- Graph meets x-axis when y 0, i.e. (-00, 0), (-0, 0), (60, 0), 40, 0) 7 a Let 7cosq + 4sinq Rcos(q -a) Rcosq cosa + Rsinq sina R > 0, 0 < a < p Compare cosq : Rcosa 7 () Compare sinq : Rsina 4 () Divide () y () : tana 4 7 Þ a.9 ( d.p.) R 4 + 7 Þ R So 7cos q + 4sin q cos(q -.9) 4cosq + 48sinq cosq + cosq ö 4 + 4(sinq cosq) 7(+ cosq) + 4sinq 7 + 7cosq + 4sinq The maximum value of 7cosq + 4sinq is ( using (a) with cos(q -.9) ) So maximum value of 7 + 7cosq + 4sinq 7 + Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
7 c Using the answer to part a: Solve cos(.9). cos(.9).9, 0.990...,.67099... 6.9, 66.9 8 a Let.sinx + cosx Rsin(x +a) Rsinxcosa + Rcosxsina R > 0, 0 < a < p Compare sinx : Rcosa. () Compare cosx : Rsina () Divide () y () : tana 4 Þ a 0.97 ( d.p.) R +. Þ R. sin xcos x + 4cos x + cosx ö (sin xcos x) + 4 sinx + + cosx sinx + cosx + c From part (a).sinx + cosx.sin(x + 0.97) So maximum value of.sinx + cosx.. So maximum value of sin xcos x + 4cos x. + 4. 9 a sin sin - cosq sinq - cosq 4sinq 4sinq + cosq Let 4sin cos Rsin( ) Rsin cos Rcos sin So Rcos 4 and Rsin Rsina Rcosa tana 4 ö a tan - 4 tan- 0. 4.04 ( d.p.) R 4 + 7 4sinq + cosq 7 sin(q +4.04 ) 7 sin(q +4.04 ), for 0 q 60 sin(q +4.04 ) 7 0.4 ( d.p.) q +4.04 sin - 0.4 4.04, for 4.04 q +4.04 74.04 q + 4.04 4.04,6.96, 74.04 q 0,.9, 60 0 a cosq + sinq So cosq - sinq Let cosq - sinq Rcos(q +a) Rcos cos Rsin sin So Rcosa and Rsina Rsina Rcosa tana ö a tan - 6. ( d.p.) R + R So cosq - sinq cos(q + 6. ) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
0 cos(q + 6. ), for 0 q 60 cos(q + 6. ), for 6. q + 6. 46. q + 6. 7.9, 86. ( d.p.) q 7.6, 9.8 ( d.p.) a LHS cosq sinq sinq cosecq RHS sinq LHS tan p 4 + tan x - tan p 4 tan x - tan p 4 - tan x + tan p 4 tan x + tan x - tan x - - tan x + tan x ( + tan x) - - tan x - tan x ( + tan x) + tan x + tan x - tan x - - tan x + tan x - tan x 4tan x - tan x tan x ö - tan x tanx RHS c LHS (sin xcos y + cos xsin y) (sin xcos y - cos xsin y) sin xcos y - cos xsin y (- cos x)cos y - cos x(- cos y) cos y - cos xcos y - cos x + cos xcos y cos y - cos x RHS d LHS + cosq + (cos q -) a LHS tan x cosq + cos q cosq(+ cosq) cosq(cos q) 4cos q cosq RHS - cosx + cosx - (- sin x) + (cos x -) sin x cos x tan x RHS tan x, for x tan x x, tan x x, x,,, a LHS cos 4 q - sin 4 q cos q - sin q ( cos q + sin q ) cos q - sin q cos4q RHS cos4q, for 0 4q 70 4q 60, 00, 40, 660 q, 7,0,6 4 a LHS - (- sin q) sinq cosq sin q sinq cosq sinq tanq RHS cosq When 80, sinq sin60 0 and cos60 0 therefore 80 is a solution of the equation sin cos Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
4 c Rearrange sin cos to give ( cos ) sin Using the identity in part (a) gives tan Þ tanq, for 0 < q < 60 q 6.6, 06.6 ( d.p.) a Set cos x sin x R cos( x ) Rcos xcosa - Rsin xsina So Rcosa and Rsina Rsina Rcosa tana ö a tan - 0.84 ( d.p.) R + 9 R cos x - sin x cos(x + 0.84) cos(x + 0.84) -, for 0.84 x + 0.84< p + 0.84 cos(x + 0.84) - x + 0.84.9, 4.7 x.07,. ( d.p.) 6 a Set.4sin.6cos Rsin( ) Rsinq cosa - Rcosq sina So Rcos.4 and Rsin.6 Rsina.6 tana Rcosa.4 a tan - 4 7.964 ( d.p.) R.4 +.6. R.77 ( d.p.) The maximum value of.77sin( 7.964) is when sin( 7.964). So the maximum value is.77 and it occurs when 7.964 90, 6.964 c -.6cos 60t ö 6 +.4sin 60t ö 6 +.77sin 60t 6-7.964 ö The minimum numer of daylight hours is when sin 60t 6-7.964 ö - So minimum is -.77 6.8 hours d sin 60t 6-7.964 ö - 60t - 7.964 70 6 t days 7 a Let sin x +cos x Rsin(x +a) Rsinxcosa + Rcos xsina So Rcos and Rsin Rsina Rcosa tana ö a tan -.6 ( d.p.) R + 69 R So sin x +cos x sin(x +.6 ) 0 v(x) sin x ö + cos x ö 0 sin x +.6 ö The minimum value of v is when x sin.6 So 0.8 m/s ( d.p.) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 8
7 c sin x +.6 ö, for.6 x +.6 66.6 x +.6 90 x 68. minutes a As ÐOAB ÐOBAÞ ÐAOB p - q, so ÐBOD q Challenge a Using cos P + cosq cos P + Q ö cos P - Q ö and sin P - sinq cos P + Q ö sin P - Q ö cosq + cos4q LHS sinq - sin4q cos 6q ö cos q ö cos 6q ö -q ö sin cosq cosq cosq sin -q cosq sin( -q ) -cotq LHS cos x + cosx + cosx cosx + cos x + cosx cos 6x ö 4x ö cos + cosx cosxcosx + cosx cosx(cos x + ) cosx(cos x) 4cos xcosx RHS OB OD cosq BD sinq AB cosq sinq BD AB BD cosq So BD sinq cosq But BD sinq So sinq sinq cosq AB cosq AD (cosq)cosq cos q OD cos q - From part (a) OD cosq So cosq cos q - Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 9